# The polarization identity in Hilbert space

1. Apr 16, 2012

### AxiomOfChoice

If we assume the inner product is linear in the second argument, the polarization identity reads

$$(x,y) = \frac 14 \| x + y \|^2 - \frac 14 \| x - y \|^2 - \frac i4 \|x + iy\|^2 + \frac i4 \| x - iy \|^2.$$

But there is another identity that I've seen referred to in some texts as the "polarization identity", and it is as follows: If T is a linear operator on the Hilbert space, we have

$$(x,Ty) = \frac 14 (x+y,T(x+y)) - \frac 14 (x-y,T(x-y)) - \frac i4 (x+iy,T(x+iy)) + \frac i4 (x-iy,T(x-iy)).$$

I've expanded this out and checked to make sure it's correct, and it is...but is there a quick and painless way to prove this by starting from the first polarization identity above? If so, I really don't see it...my professor seemed to indicate to us that if we defined a kind of ersatz "inner product" on H by letting $(x,y)_1 = (x,Ty)$, then the result would follow if we wrote out the first polarization identity with the norm $\| \cdot \|_1$ instead. But is this really kosher?

Also, just out of curiosity...why is it appropriate to call this thing a "polarization" identity?

2. Apr 16, 2012

### mathman

I haven't worked out the details myself. However if you work out the squares of each of the terms in the first equation, all the square terms cancel and you will be left with the inner product terms, which should reproduce the second equation or some variant.

3. Apr 16, 2012

### micromass

Your professor's method is not really kosher. The problem is that $(x,Ty)$ does not define an inner product in general, but only if T is hermitian and positive semi-definite.

Of course, the polarization identity can (probably) be proven in the general case that (x,y) is a conjugate-bilinear map instead of an inner product. But you must really check that you did not use the other properties of an inner product.